The function `f(x)=int_(0)^(x) log _(|sin t|)(sin t + (1)/(2)) dt`, where `x in (0,2pi)`, then `f(x)` strictly increases in the interval

Options

A. `((pi)/(6),(5pi)/(6))`
B. `((5pi)/(6),2pi)`
C. `((pi)/(6),(7pi)/(6))`
D. `((5pi)/(6),(7pi)/(6))`

Answer

To determine the intervals in which the function \( f(x) = \int_{0}^{x} \log_{|\sin t|} \left( \sin t + \frac{1}{2} \right) dt \) is strictly increasing, we will follow these steps:

### Step 1: Find the derivative \( f'(x) \)

Using the Fundamental Theorem of Calculus, we can differentiate \( f(x) \):

\[ f'(x) = \log_{|\sin x|} \left( \sin x + \frac{1}{2} \right) \]

### Step 2: Determine when \( f'(x) > 0 \)

For \( f(x) \) to be strictly increasing, we need:

\[ f'(x) > 0 \]

This means:

\[ \log_{|\sin x|} \left( \sin x + \frac{1}{2} \right) > 0 \]

### Step 3: Analyze the logarithmic condition

The logarithm is positive when its argument is greater than 1. Thus, we need:

\[ \sin x + \frac{1}{2} > 1 \]

This simplifies to:

\[ \sin x > \frac{1}{2} \]

### Step 4: Analyze the base condition

For the logarithm to be defined, the base must also be positive:

\[ |\sin x| > 0 \]

This implies:

\[ \sin x > 0 \]

### Step 5: Solve the inequalities

1. From \( \sin x > \frac{1}{2} \):
- The solutions are in the intervals:
\[ x \in \left( \frac{\pi}{6}, \frac{5\pi}{6} \right) \]

2. From \( \sin x > 0 \):
- The solutions are in the intervals:
\[ x \in (0, \pi) \]

### Step 6: Find the intersection of the intervals

Now, we need to find the intersection of the intervals \( \left( \frac{\pi}{6}, \frac{5\pi}{6} \right) \) and \( (0, \pi) \):

The intersection is:

\[ x \in \left( \frac{\pi}{6}, \frac{5\pi}{6} \right) \]

### Conclusion

Thus, the function \( f(x) \) is strictly increasing in the interval:

\[ \left( \frac{\pi}{6}, \frac{5\pi}{6} \right) \]